A298409 a(n) = 2*a(n-1) - a(n-3) + 2*a(floor(n/2)) + 3*a(floor(n/3)) + ... + n*a(floor(n/n)), where a(0) = 1, a(1) = 2, a(2) = 3.

1, 2, 3, 15, 48, 123, 300, 635, 1316, 2555, 4873, 8850, 16096, 28296, 49424, 84749, 144733, 243607, 409156, 680308, 1128889, 1861633, 3063978, 5020133, 8217296, 13409702, 21862824, 35575784, 57853195, 93954953, 152524643, 247386674, 401132014, 650065133

Offset: 0

Author

Clark Kimberling, Feb 10 2018

Keywords

• nonn
• easy

Comments

a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio (A001622), so that (a(n)) has the growth rate of the Fibonacci numbers (A000045). See A298338 for a guide to related sequences.

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A001622, A000045, A298338, A298408.

Programs

Mathematica

a[0] = 1; a[1] = 2; a[2] = 3;
a[n_] := a[n] = 2*a[n - 1] - a[n - 3] + Sum[k*a[Floor[n/k]], {k, 2, n}];
Table[a[n], {n, 0, 90}]  (* A298409  *)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A298409(n):
    if n <= 2:
        return n+1
    c, j = 2*A298409(n-1)-A298409(n-3), 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2*(j2-1)-j*(j-1))*A298409(k1)//2
        j, k1 = j2, n//j2
    return c+2*(n*(n+1)-j*(j-1))//2 # Chai Wah Wu, Mar 31 2021